Problem: Simplify the following expression and state the condition under which the simplification is valid. $x = \dfrac{-8r^2 - 16r + 24}{4r^3 - 20r^2 - 96r}$
Solution: First factor out the greatest common factors in the numerator and in the denominator. $ x = \dfrac {-8(r^2 + 2r - 3)} {4r(r^2 - 5r - 24)} $ $ x = -\dfrac{8}{4r} \cdot \dfrac{r^2 + 2r - 3}{r^2 - 5r - 24} $ Simplify: $ x = - \dfrac{2}{r} \cdot \dfrac{r^2 + 2r - 3}{r^2 - 5r - 24}$ Next factor the numerator and denominator. $ x = - \dfrac{2}{r} \cdot \dfrac{(r + 3)(r - 1)}{(r + 3)(r - 8)}$ Assuming $r \neq -3$ , we can cancel the $r + 3$ $ x = - \dfrac{2}{r} \cdot \dfrac{r - 1}{r - 8}$ Therefore: $ x = \dfrac{ -2(r - 1)}{ r(r - 8)}$, $r \neq -3$